Grinstead-Snell-Probability · Problem 18

Expected Value and Variance

Mathematics · variance-discrete-random-variables

Problem

Show that, for the sample mean

\bar{x}

and sample variance

s^2

as defined in Exercise 17, ((a))

E(\bar{x}) = \mu

. ((b))

E\left((\bar{x} - \mu)^2\right) = \sigma^2/n

. ((c))

E(s^2) = \frac{n-1}{n}\sigma^2

. Hint: For (c) write

\sum_{i=1}^n (x_i - \bar{x})^2 = \sum_{i=1}^n \left((x_i - \mu) - (\bar{x} - \mu)\right)^2 = \sum_{i=1}^n (x_i - \mu)^2 - n(\bar{x} - \mu)^2,

and take expectations of both sides, using part (b) when necessary. ((d)) Show that if, in the definition of

s^2

in Exercise 17, we replace the coefficient

1/n

by the coefficient

1/(n-1)

, then

E(s^2) = \sigma^2

. (This shows why many statisticians use the coefficient

1/(n-1)

. The number

s^2

is used to estimate the unknown quantity

\sigma^2

. If an estimator has an average value which equals the quantity being estimated, then the estimator is said to be unbiased. Thus, the statement

E(s^2) = \sigma^2

says that

s^2

is an unbiased estimator of

\sigma^2